y’’+4y’+3y=0 y(0)=2 y’(0)=-1
r^2+4r+3=0
(r+3)(r+1)=0
r=-3 r=-1
y(t)=c1e^-3t+c2e^-t
Since y(0)=2
y(0)=c1+c2=2
c2=2-c1 (1)
y’(t)=-3c1e^-3t-c2e^-t
y’(0)=-1
y’(0)=-3c1-c2=-1
-3c1-(2-c1)=-1
-3c1-2+c1=-1
-2c1=1
c1=-1/2
c2=5\2 from(1)
y(t)=-1\2e^-3t+5\2e^-t
